The expansion of (1+x)^4 is 1 + 4x +nx^2 + 4x^3 + x^4. Find the value of n. Hence Find the integral of (1+√y)^4 between the values 1 and 0 (one top, zero bottom).

Using Binomial expansion or Pascal's triangle, expand (1+x)^4 to get 1+4x+6x^2+4x^3+x^4. Then, by substituting √y for x, get 1 + 4y^1/2 + 6y +4y^3/2 +y^2. Then, using the rules of integration, the expansion is integrated to y + 8/3y^3/2 + 3y^2 + 8/5y^5/2 + 1/3y^3 between the bounds 1,0. substituting in the values gives [1 + 8/3 + 3 + 8/5 + 1/3] - = 7 + 8/5 = 8.6.